Binomial Distribution: Definition, Properties and Solved Examples

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, each with the same probability of success. It is defined by two parameters: the number of trials (n) and the probability of success in each trial (p). The formula to calculate the probability of exactly k successes is P(X=k) = (ⁿ_k)p^k(1−p)^n-k. Key properties include mean (np) and variance (npq, where q = 1 – p). This distribution is fundamental for analyzing binary outcomes and is frequently tested in competitive exams like IIT-JEE, GATE, GRE, and CAT, where solving related problems helps understand real-life applications like quality control and decision-making under uncertainty.

What is Binomial Distribution?

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent and identical trials, where each trial has only two possible outcomes: success or failure. The probability of success remains constant across trials, and each trial is independent of others. The distribution is used to calculate the likelihood of achieving a specific number of successes in a given set of trials. Important terms related to binomial distribution are:

1. Trial: A single event with two possible outcomes (success/failure).
2. Success (p): The probability of the desired outcome in a single trial.
3. Failure (q): The probability of the undesired outcome, where q=1−p.
4. Number of Trials (n): The total number of independent trials or experiments conducted.
5. Number of Successes (k): The number of successful outcomes in the trials.
6. Binomial Coefficient (ⁿ_k): A combination that calculates the number of ways to choose k successes from n trials, represented as n!/k!(n−k)!​.
7. Probability Mass Function (PMF): The formula used to calculate the probability of exactly k successes: P(X=k) = (ⁿ_k)p^k(1−p)^n-k.
8. Mean (np): The expected number of successes in n trials.
9. Variance (npq): A measure of the spread or variability in the number of successes.

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Properties of Binomial Distribution

The binomial distribution has several key properties that describe its behavior and characteristics. These properties are crucial in understanding the distribution’s shape, central tendency, and variability

1. Two Possible Outcomes: Each trial has only two possible outcomes: success (with probability p) and failure (with probability q=1−p).
2. Fixed Number of Trials (n): The number of trials, n, is fixed and predetermined. Each trial is independent of others, meaning the outcome of one trial does not affect the others.
3. Constant Probability (p): The probability of success, p, remains the same for all trials.
4. Discrete Distribution: The binomial distribution is discrete, meaning it takes only integer values (the number of successes) ranging from 0 to n.
5. Mean (Expected Value): The mean or expected value of the binomial distribution is given by:
   μ = np
   This represents the average number of successes expected in n trials.
6. Variance: The variance, which measures the spread of the distribution, is:
   σ² = npq
   where q = 1−p is the probability of failure. This shows how much the number of successes is expected to deviate from the mean.
7. Standard Deviation: The standard deviation is the square root of the variance: σ = √npq
8. Symmetry and Skewness: The binomial distribution becomes more symmetric as n increases and when p is close to 0.5. If p is much greater than 0.5, the distribution is skewed to the left (negatively skewed), while if p is much less than 0.5, it is skewed to the right (positively skewed).
9. Maximum Probability: The most probable value (the mode) of the binomial distribution is around k = np (the mean), but it may vary slightly depending on whether p is closer to 0 or 1.
10. Relationship with Normal Distribution: For large n, the binomial distribution approaches a normal distribution, especially when p is close to 0.5. This is known as the normal approximation to the binomial distribution, often applied when np ≥ 5 and nq ≥ 5.

Formula of Binomial Distribution

The formula of the binomial distribution gives the probability of obtaining exactly k successes in n independent trials, where each trial has two possible outcomes (success or failure) and the probability of success is p.

Binomial Distribution Formula:

P(X=k) = (ⁿ_k)p^k(1−p)^n-k

Where:

• P(X=k): The probability of getting exactly k successes in n trials.
• n: The total number of trials.
• k: The number of successes.
• p: The probability of success in a single trial.
• 1−p or q: The probability of failure in a single trial.
• (ⁿ_k): The binomial coefficient, representing the number of ways to choose k successes from n trials, is calculated as: 

(ⁿ_k) = n!/k!(n−k)!​ 

where n! is the factorial of n.

Explanation:

• The term p^k represents the probability of getting k successes.
• The term (1−p)^n-k represents the probability of getting n – k failures.
• The binomial coefficient (ⁿ_k) accounts for the different combinations in which the k successes can occur in n trials.

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Solved Examples of Binomial Distribution

Here are 5 solved examples of binomial distribution to illustrate its application:

Example 1: Tossing a Coin

You toss a fair coin 5 times. What is the probability of getting exactly 3 heads?

• Solution:
  • Number of trials n = 5
  • Probability of success (head) p=0.5
  • Probability of failure (tail) q = 1−p = 0.5
  • Number of successes k = 3

Using the binomial distribution formula:

P(X=3) = (⁵₃)(0.5)³(0.5)^5-3 

P(X=3) = {5!/3!(5−3)!}(0.5)⁵ = 10×0.03125 = 0.3125

So, the probability of getting exactly 3 heads is 0.3125.

Example 2: Quality Control

A factory produces light bulbs, and 95% of them are good. If you randomly select 10 bulbs, what is the probability that exactly 9 of them are good?

• Solution:
  • Number of trials n = 10
  • Probability of success (good bulb) p = 0.95
  • Probability of failure (bad bulb) q = 0.05
  • Number of successes k = 9

Using the formula:

P(X=9) = (¹⁰₉)(0.95)⁹(0.05)¹ 

P(X=9) = 10×(0.95)9×(0.05)

P(X=9) ≈ 10×0.6302×0.05 = 0.3151

So, the probability of exactly 9 good bulbs is 0.3151.

Example 3: Exam Questions

In a multiple-choice exam, each question has 4 options, with only one correct answer. If you randomly guess on 6 questions, what is the probability of getting exactly 2 correct answers?

• Solution:
  • Number of trials n = 6
  • Probability of success (correct answer) p = ¼ = 0.25
  • Probability of failure (wrong answer) q = 0.75
  • Number of successes k = 2

Using the formula:

P(X=2) = (⁶₂)(0.25)²(0.75)⁴ 

P(X=2) = {6!/2!(6−2)!}×(0.25)²×(0.75)⁴ = 15×0.0625×0.3164P 

P(X=2) ≈ 15×0.01977 = 0.2966

The probability of getting exactly 2 correct answers is 0.2966.

Example 4: Defective Products

A shipment contains 12 items, and there is a 20% chance that any item is defective. If you randomly check 8 items, what is the probability that exactly 2 are defective?

• Solution:
  • Number of trials n = 8
  • Probability of success (defective item) p = 0.2
  • Probability of failure (non-defective) q = 0.8
  • Number of successes k = 2

Using the formula:

P(X=2) = (⁸₂)(0.2)²(0.8)⁶ 

P(X=2) = 28×(0.2)²×(0.8)⁶ = 28×0.04×0.2621 

P(X=2) ≈ 28×0.010484 = 0.2936

The probability of finding exactly 2 defective items is 0.2936.

Example 5: Drug Trial

In a drug trial, a new treatment has a 70% success rate. If 10 patients are treated, what is the probability that exactly 7 patients will recover?

• Solution:
  • Number of trials n = 10
  • Probability of success (recovery) p = 0.7
  • Probability of failure (no recovery) q = 0.3
  • Number of successes k = 7

Using the formula:

P(X=7) = (¹⁰₇)(0.7)⁷(0.3)³ 

P(X=7) = {10!7!(10−7)!}×(0.7)⁷×(0.3)³ = 120×0.0823543×0.027 

P(X=7) ≈ 120×0.002224 = 0.2669

The probability of exactly 7 patients recovering is 0.2669.

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